120132+320134+520136 + . . . +992013100

1–2+3–4+5–6 + . . . +99–100

 

Answer:

-50

Step-by-step explanation:

This was a classic fave of our math teacher. You can answer this by treating the question as two separate arithmetic series.

The first one would be the odd numbers:

1 + 3 + 5 + ... + 99, with a common difference of 2.

The other one would be the even numbers:

-2 - 4 - 6 - ... - 100, with a common difference of -2.

Solve the series as you would any other.

Sn = (a₁ + aₙ) (n/2)

First, we need to find the number of terms for each. You can solve this using the same formula above, but since we know that it counts to 100, there are 50 odd and 50 even terms.

odd:

S₅₀ = 25(a₁ + a₅₀)

S₅₀ = 25(1 + 99)

S₅₀ = 25(100)

S₅₀ = 2500

even:

S₅₀ = 25(a₁ + a₅₀)

S₅₀ = 25(-2 - 100)

S₅₀ = 25(-102)

S₅₀ = -2550

The total of both arithmetic series would then be -50.